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Codeforces Round #267 (Div. 2) C

2024-07-23 08:04:13   218人阅读

题目：

C. George and Job
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output

The new ITone 6 has been released recently and George got really keen to buy it. Unfortunately, he didn‘t have enough money, so George was going to work as a programmer. Now he faced the following problem at the work.

Given a sequence of n integers p1,?p2,?...,?pn. You are to choose k pairs of integers:

[l1,?r1],?[l2,?r2],?...,?[lk,?rk] (1?≤?l1?≤?r1?<?l2?≤?r2?<?...?<?lk?≤?rk?≤?n; ri?-?li?+?1?=?m),?

in such a way that the value of sum  is maximal possible. Help George to cope with the task.

Input

The first line contains three integers n, m and k (1?≤?(m?×?k)?≤?n?≤?5000). The second line contains n integers p1,?p2,?...,?pn (0?≤?pi?≤?109).

Output

Print an integer in a single line — the maximum possible value of sum.

Sample test(s)
input
5 2 1
1 2 3 4 5
output
9
input
7 1 3
2 10 7 18 5 33 0
output
61

题意分析：

DP一下。dp[I][j]存前i有j组的最大值。

状态转移方程：
if(i-m>=0)
	dp[i][j] = max(dp[i][j],max(dp[i-1][j],dp[i-m][j-1]+b[i]-b[i-m])) ;
else 
<span style="white-space:pre">	</span>dp[i][j] = max(dp[i][j],dp[i-1][j]) ;
其中b[i]存前i项a[i]的和。

代码:

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>


using namespace std;

long long dp[5005][5005] ;
long long a[65555] ;
long long b[65555] ;
int main()
{
	int n,m,k ;
	cin>>n>>m>>k;
	int i,j;
	for(i=1;i<=n;i++)
	{
		cin>>a[i];
		b[i] = b[i-1] + a[i] ;
	}
	for(i=1;i<=n;i++)
	{
		for(j=1;j<=k;j++)
		{
			if(i-m>=0)
			dp[i][j] = max(dp[i][j],max(dp[i-1][j],
				dp[i-m][j-1]+b[i]-b[i-m])) ;
			else dp[i][j] = max(dp[i][j],dp[i-1][j]) ;
		}
	}
	cout<<dp[n][k]<<endl;
	return 0 ;
}



Codeforces Round #267 (Div. 2) C


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